Question

A man pulls a suitcase of mass 15 kg with a force "F" inclined at an angle of 60° to the horizontal. The frictional force between the floor and the wheels of the suitcase is 10 N. What is the force required to produce an acceleration of 0.1 m/s2 on the suitcase?

Answer 1

Answer:

F=23N

Explanation:

The net force applied on the suitcase is the sum of the force F the man applies, the friction, the weight and the normal from the ground. We need only to look at the horizontal component of the forces to solve the question asked. On the horizontal, the net force would be the horizontal component of F minus the frictional force f, which always opposes movement (all the other forces are on the vertical component). So we have:

$F_{Nx}=F_x-f$

We now apply Newton's second Law on the horizontal:

$F_{Nx}=ma_x$

And we calculate the horizontal component of F as $F_x=Fcos(60)=0.5F$, so we have

$ma_x=F_x-f=0.5F-f$

And to solve for F we do:

$ma_x+f=0.5F$

$F=\frac{ma_x+f}{0.5}$

And substitute our values:

$F=\frac{(15kg)(0.1m/s^2)+10N}{0.5}=23N$